Universal space for the family of subgroups of a finite cyclic group

Question

Let $G$ be a compact Lie group and let $\mathcal{P}_G$ denote the family of proper subgroups of $G$. The universal space for the family $\mathcal{P}_G$ is a cofibrant $G$-space which does not have $G$-fixed points and such that for every proper subgroup $H<G$, the fixed point space $(E\mathcal{P}_G)^H$ is contractible. These properties determine $E\mathcal{P}_G$ uniquely up to $G$-homotopy equivalence.

I am particularly interested in the case where $G$ is a finite cyclic group. When the order of $G$ is prime, I know that I can take the model $E\mathcal{P}_G=EG$, the non-equivariant universal space associated to the group $G$. What I would like to know is if there are concrete models for $E\mathcal{P}_G$ for general cyclic groups. My end goal is to compute the mod-2 singular homology of such spaces, but I find it difficult to do so with just the properties characterising such spaces.

In general, I am also interested in knowing how one goes about constructing $E\mathcal{P}_G$ for general (compact Lie) groups.

Answer 1

Let $X$ be a space with an action of $G$ such that $X^G=\emptyset$, and for every proper subgroup $H$, $X^H\ne \emptyset$. Then the infinite join of $X$ with itself is a universal space for the family of proper subgroups. This follows from the following facts: fixed points commute with join, infinite join of non-empty spaces is contractible, infinite join of empty spaces is empty.

So for example, if $G={\mathbb Z}/p^n$ is a cyclic group of prime power order, you can let $G$ act on $S^1$ via the standard action of the quotient group $G/pG\cong {\mathbb Z}/p$ on the circle. This action has the required property, so the infinite join of this $S^1$ with itself is a model for the universal space. It is homeomorphic to $S^\infty$. Added later: It is easy to see that this is just $E{\mathbb Z}/p$, which is acted on by $G$ via the quotient map. You say that your eventual goal is to compute the mod-$2$ homology of such spaces. The universal space itself is contractible. Perhaps you are interested in the homology of the quotient space $EP_G/G$? In the case $G={\mathbb Z}/p^n$, the orbit space is simply the classifying space $B{\mathbb Z}/p$, whose homology is well-known.

A general finite cyclic group of order $n$ is a product of prime power cyclic groups, with distinct primes. You can take the join of universal spaces of the factors to get a universal space for the whole group. The orbit space is the join of spaces $B{\mathbb Z}/p$, where $p$ ranges over the prime factors of $n$. So its mod $2$ homology is the same as of ${\mathbb RP^\infty}$ if $n$ is a power of $2$, and is the same as of a point if $n$ has an odd factor.